Post by: Eternal Student on 17/04/2023 00:51:18
Hopefully the poll speaks for itself. The forum looked quiet this evening and this might provide some discussion or activity.
There are some answers for this already. You don't need to go and look for them in textbooks or on the internet but you can if you had a few minutes to spare. As always, use your own discretion and try to select a sensible and safe website.
Best Wishes.
Post by: Halc on 17/04/2023 03:29:25
There is a well known model in Quantum mechanics called "the particle in a box" (also "a particle in an infinite square well"). Considering that model, does the particle exert pressure on the walls of the box?I had to look up exactly what these words meant. "Box" and "Infinite square well" sort of imply a large container, where in fact they're talking about a very small one in a deep potential well from which escape (by tunneling say) isn't possible.
Without reading any bit about pressure, I'd say yes, it would since it could occupy a lower energy state if it had more room for a longer wavelength. It can exist only in certain quantized energy states in there, and a wider box allows a lower energy one. So it applies pressure the same way that water would since water could occupy a lower energy state with more leg room.
Post by: Eternal Student on 17/04/2023 03:56:19
That's a fair answer. I need to lay down some opposition just to keep the poll interesting. (Just to be clear, it doesn't matter what I think, we just need arguments on both sides).
One important feature of the model is that the wave function of the particle looks like this:
(https://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/InfiniteSquareWellAnimation.gif/200px-InfiniteSquareWellAnimation.gif)
Diagram A is NOT the quantum model, it is the classical Newtonian understanding of what should be happening. In that model, the particle is sometimes at the walls and can exchange momentum there.
Diagrams B through to F illustrate the wave function representation which is part of the quantum mechanical model of a particle in a box. One of the important points about this model is that the wave function MUST always be zero at the walls of the box. The modulus squared of that wave function (or in simpler English "the size" of it) tells you the probability of finding the particle at that location. So the particle is NEVER at the walls of the box. So it cannot impart momentum to those walls.
Best Wishes.
Post by: Eternal Student on 17/04/2023 04:29:54
We might need another counter-argument.
Just considering the 1-dimensional situation for simplicity (like the diamgrams in the previous post): If you measure a force on a wall at one moment of time, then you know precisely where the particle was and the momentum that it had at that time. (It was bouncing off the wall). This would violate the uncertainty principle. You cannot know the precise location and momentum of the quantum particle simultaneously. The only sensible escape is that it cannot happen - you cannot measure a force on a wall at any moment in time.
Best Wishes.
Post by: Halc on 17/04/2023 13:15:44
So the particle is NEVER at the walls of the box. So it cannot impart momentum to those walls.Does it say that somewhere? Because this seems a non sequitur to me.
First of all, the particle cannot be found at the walls. Wrong to say it is somewhere. But I'm talking about the conclusion that it cannot impart pressure (not momentum) to the walls just because it never gets measured right there.
Post by: Eternal Student on 17/04/2023 13:52:10
Does it say that somewhere?Where does it state that the particle does or can exert force on the walls? Ideas like obtaining a force are macroscopic interpretations. Quantum Mechanics for the particle in the box consists of just a handfull of postulates. They can be written in a few different ways, the wording is certainly flexible, but here is one version:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html
(It's the second boxed section that lists 6 postulates. The last box just expands on what is meant by items 1. and 5. and provides precisely the result we needed to determine the probability of finding the particle somewhere).
If you wish to assert that QM does directly permit particles to transmit a force to something else (like the walls) then you need to construct or exhibit a suitable Hermitian operator that represents that force (or provide some reference to where that is done).
Wrong to say it is somewhere.That is actually fair enough, it is only true that it won't be found at the walls when you measure its position. However what constitutes a measurement? If you did observe an impulse at one wall for one brief moment of time, doesn't that constitute a measurement of the position of the particle? See the discussion earlier about the uncertainty principle. This simple QM model does not include multiple particles: It is not as if the particle could have ejected a photon from a remote location to impact on the wall.
Final Note: I'm not stating my own views, just deliberately presenting arguments for the other side.
Best Wishes.
Post by: Eternal Student on 17/04/2023 13:58:43
I've had one response to the poll so far and that was me.
* It is open to guests, you do not need to have created an account.
* This is in the Just Chat section and not serious. It's also anonymous (the poll is anonymous but written replies will include your username).
Best Wishes.
Post by: Halc on 17/04/2023 16:08:11
Where does it state that the particle does or can exert force on the walls?That's what is under discussion I thought. I'm not asserting anything.
Quote
I'm not stating my own views, just deliberately presenting arguments for the other side.In that spirt, yes.
I just said that the zero probablilty of finding the particle right at the wall does not necessitate that no pressure is applied to the walls. It's not even true in classical physics where some atom gets close to the wall but never actually touches it, but nevertheless transfers momentum to it when it accelerates away.
Quote
Ideas like obtaining a force are macroscopic interpretations.Agree, but you get macroscopic effects from many quantum effects. So you might measure a pressure of sorts, but it's probably wrong to suggest that a super-sensitive meter would register spike accelerations of the wall on a periodic basis like you would with a red ball bouncing back and forth.
I based my pressure on the naive suggestion that the tight walls necessarily put the particle in a higher minimal energy state. It has potential energy, a bit like a compressed spring.
Quote
If you did observe an impulse at one wall for one brief moment of time, doesn't that constitute a measurement of the position of the particle?I imagine it would, but I'm not suggesting a periodic impulse any more than a nitrogen tank under high pressure exhibits back&forth motion as the particles collide with the container. The pressure would be balanced, exerted on both sides, as would the pressure from said compressed spring. Again, my naive guess. I'm hardly an authority here.
Post by: Eternal Student on 17/04/2023 21:11:24
Classical results should be recoverable from QM when you take large numbers of particles and work only with averages or expected values. How does your situation match up with the temperature dependency that we do observe for a gas?
This formula:
Energy (of state n) = En =
with n = energy level, L = length of one side (assuming the box has all sides the same length for 2-D or 3-D), can be derived from the QM model. One important feature is that n is a strictly positive integer n =1, 2, 3, .... we cannot have n=0 or else the wave function becomes 0 and the particle has disappeared. So there is a minimum energy or ground state. The best you can hope for is that lowering temperature tends to put the particle in a lower energy state but eventually it hits the minimum energy state. This is unavoidable.
However, that energy would always vary ~ 1/L2 so that, by your argument, the particle can always obtain a lower energy by expanding the space available to it. It will always exert some pressure on the walls of the box no matter how cold you make it. How do you reconcile that with what we know about ideal gases?
Best Wishes.
Post by: alancalverd on 18/04/2023 09:32:06
This aligns with Halc's explanation. If the box were wider, the energy density inside it would be lower, so there is pressure on the walls. However Planck stuck with classical mechanics to the extent that his walls were perfectly rigid and elastic, so no work is done on reflection and the particle retains all its energy.
That also explains the nodes at the walls. The particle bounces off in an infinitesimal time (because no energy is transferred) so the time it resides in the vicinity of the wall, and thus the probability of finding it there, tends to zero.
Post by: alancalverd on 18/04/2023 09:36:12
How do you reconcile that with what we know about ideal gases?We don't know anything about ideal gases, because there aren't any! The classical ideal gas consists of particles with mass but no radius, so it is infinitely compressible.
Post by: Eternal Student on 18/04/2023 16:49:47
Worth considering Einstein's equivalence of energy density with pressure.I'm not aware of a statement of such an equivalence principle. However, there is almost an equivalence in General Relativity, if that's what you meant.
The stress-energy tensor of GR includes components due to the pressure so that pressure could influence curvature in a way broadly similar to the distribution of energy through space.
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/StressEnergyTensor_contravariant.svg/236px-StressEnergyTensor_contravariant.svg.png)
Figure 1: The stress-energy tensor is shown with pressure components in green, energy density in red and related energy flux in yellow.
However, it is not clear that you could obtain an identical curvature tensor from the Einstein Field Equation of GR by replacing all energy density with pressure, even allowing for a change of co-ordinates to be applied to the resulting curvature tensor. More generally, I am not aware of any straight forward relationship like E=mc2 (expressing an equivalence of mass and energy) that expresses an equivalence of pressure and energy density.
If you have a better reference for this "pressure and energy equivalence" of which you speak, please let me know. It would be interesting to read.
- - I'm keeping the post short and just ending this here ---
Best Wishes.
Post by: Bored chemist on 18/04/2023 20:02:28
You run into a different conundrum.
(https://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/InfiniteSquareWellAnimation.gif/200px-InfiniteSquareWellAnimation.gif)
In the first excited state- labelled C- your particle has zero chance of being in the middle of the box.
But it has a 50% chance of being one either side.
So, how does it get from left to right without ever being in the middle?
You can pretend it doesn't matter- it's actually really on one side of the box and, unless it's kicked into a higher state, it remains there.
But it's worse than that.
The observable universe can be treated as a huge box with very high walls.
Because the box is big, the energy level separation is tiny.
So any particle you see will actually be in a very high excited state.
It will be passing through a massive number of the "zeroes" of probability density every second.
How does any particle travel past the bits where its probability is zero?
Post by: Bored chemist on 18/04/2023 20:06:59
We don't know anything about ideal gases, because there aren't any!So... that's one thing we know about them.
I think we know other stuff about them too.
They are colourless. (If they interacted with photons they would interact with those emitted by the walls of the container (and one-another) and that's forbidden by their definition.)
Arguably, we know everything about them for the same reason that Tolkien knew everything about Hobbits.
Post by: alancalverd on 19/04/2023 08:25:44
Post by: alancalverd on 19/04/2023 08:33:58
I'm not aware of a statement of such an equivalence principle.I've only come across its ascription to Einstein in a few lectures and can't find it on line, but I'm sure it's in at least one of his publications. Anyway the story is
Dimensions of energy: ML2T-2
Dimensions of volume: L3
Dimensions of pressure = force/area = MLT-2/L2 = ML-1T-2 = energy/volume
It's an extremely useful equivalence in astrophysics and mechanical engineering!
PS apologies - I just ticked the wrong box in the poll! I refer the hon gent to the answer I gave yesterday.
Post by: Eternal Student on 19/04/2023 12:53:55
You run into a different conundrum.......how does it get from left to right without ever being in the middle?
A very good set of questions or discussion points. "Quantum weirdness" is a short answer that might be given. We could have a longer discussion but there's already plenty of articles on the internet. In this particular situation it's not too difficult to try and match up some Newtonian explanation but it may be best not to even try. Anyway, that's the attitude I need for the next paragraph.
One of the points which I still think is important is that there is a jump from the Quantum Mechanical model to the notion that a pressure would exist on the walls. At the moment those who have submitted a response to the poll have all said "yes" there is a pressure rather than "no" with one possible explanation being simply that the QM model does not take you that far. No-one has exhibited an operator that represents pressure and acts on the wave function. I need to make it clear again that without a suitable operator, "Pressure" or Force on the wall has not been established as an observable under the postulates of Quantum Mechanics. Instead all we have is a jump from the QM model to much more macroscopic properties and very Newtonian mechanics.
(Note that I am still trying to present arguments on the opposing side, no-one else is doing that at the moment).
The observable universe can be treated as a huge box with very high walls.Maybe.... The potential that a particle is being exposed to is not likely to be 0 (or some constant reference point) everywhere inside that box and I can see little reason why a particle is being kept away from the edges of the universe by some large potential (if indeed there are edges to the universe). It's a possible model but with enough assumptions and simplifications that any results can't be taken too literally.
While we're discussing the assumptions of a model, the entire "particle in a box" model is idealised and almost certainly unrealistic. We don't think there are any potentials that could produce vertical walls (change from zero to non-zero potential over 0 distance) and we don't think there are potentials that could reach an infinite value either. That is not a criticism of your attempt to model a particle in the universe. All uses of the "particle in a box" model should be viewed as just an approximation.
It's an extremely useful equivalence in astrophysics and mechanical engineering!Oddly enough dimensional analysis often does pre-empt or herald some deeper connection between two quantities - but we seem to have a small way to go yet.
Best Wishes.
Post by: alancalverd on 19/04/2023 14:34:55
One of the points which I still think is important is that there is a jump from the Quantum Mechanical model to the notion that a pressure would exist on the walls. At the moment those who have submitted a response to the poll have all said "yes" there is a pressure rather than "no" with one possible explanation being simply that the QM model does not take you that far. No-one has exhibited an operator that represents pressure and acts on the wave function. I need to make it clear again that without a suitable operator, "Pressure" or Force on the wall has not been established as an observable under the postulates of Quantum Mechanics. Instead all we have is a jump from the QM model to much more macroscopic properties and very Newtonian mechanics.Something of a cart/horse inversion, I think.
You have to remember that Planck invented quantum mechanics by derivation from a classical particle in a box model, not the other way around, by stating that the only way that the particle can conserve energy is by having nodes at the perfectly elastic and immovable walls. This then defines the permissible wave functions within the box (must have an integer number of antinodes) and the force exerted on the walls by the particle.
Since the particle is also a molecule of a perfect gas, you could derive radiation pressure from the work done to compress the gas, again a classical macroscopic process, and to nobody's surprise it delivers the experimental result for photon (the next best thing to a perfect gas) radiation pressure.
There's surely nothing weird about it. The quantum number of a particle in a one-dimensional box is the number of antinodes in its probability distribution.
Post by: alancalverd on 19/04/2023 16:48:45
Diagram A is NOT the quantum model,Oh yes it is! Everything else derives from it because it only imposes one boundary condition (nodes at the walls) and permits any number of solutions that meet that criterion, without suggesting how they could be achieved.
Post by: alancalverd on 19/04/2023 17:01:19
So, how does it get from left to right without ever being in the middle?That is indeed the difference between continuum and quantum physics. If the particle were charged, it couldn't actually oscillate because it would lose energy in doing so. The connection between continuum and quantum is only that the continuum model of a particle in a box correctly predicts the existence of nodes and antinodes of probability density, and the quantisation of energy levels.
Post by: Eternal Student on 19/04/2023 21:32:26
Some interesting posts which I have spent nearly a day thinking about.
You have to remember that Planck invented quantum mechanics by derivation from a classical particle in a box model, not the other way around....A minor but noteworthy issue is that Planck did not invent the modern version of Quantum Mechanics. It was never a single-handed development. Many of the ideas from Planck and a few others like Bohr were motivational but did not become axioms of modern quantum mechanics.
This phase (pre 1925 and involving many of Planck's ideas and Bohr's quantised model of the Hydrogen atom) is known as the old quantum theory. Never complete or self-consistent, the old quantum theory was rather a set of heuristic corrections to classical mechanics...
Building on de Broglie's approach, modern quantum mechanics was born in 1925, when the German physicists Werner Heisenberg, Max Born, and Pascual Jordan developed matrix mechanics and the Austrian physicist Erwin Schrödinger invented wave mechanics.
Extracts taken from the History of the development of Quantum Mechanics, https://en.wikipedia.org/wiki/Quantum_mechanics#History
It is perfectly reasonable to start from the postulates of modern QM (let's say as it existed post 1930) and work outwards from there. None-the-less it's interesting to consider it the other way around and your posts were well considered, thank you.
Best Wishes.
Post by: paul cotter on 20/04/2023 11:54:31
Post by: alancalverd on 20/04/2023 12:58:09
does a quantum phenomenon produce a macroscopic classical effectIt must, albeit a very small one. One test of a quantum hypothesis is whether you can derive a macroscopic observable simply by integrating lots of quantum events. A photon has no knowledge of the existence of others, so radiation pressure cannot suddenly come into being when n > 1!
Post by: Eternal Student on 20/04/2023 17:44:24
There's nothing wrong with what you've said @paul cotter . I'm grateful for anyone spending some time here and it is just chat.
The poll has nearly reached the end of its time, so I might as well say a bit more about it.
* I don't think there is a correct answer, arguments exist on both sides. However, the most conventional answer or the one which is generally accepted is "yes", the particle would exert pressure on the walls.
The general reasoning is of some small interest and easily understood. So I'll take 2 minutes and present that here.
The energy that a particle in a box can have is given by:
E =
[Equation 1]
with L = length of the box, m = mass of the particle, h = plancks constant and n = 1,2,3,4,...... The derivation of that can be found in most places discussing Quantum Mechanics. The formula holds for a 1-dimensional, 2-D or 3-D box without adjustment provided that in the higher dimensions we insist the box is a square or cube - it has the same length along all sides.
Now you just assume that the walls of the box were not fixed in place but could be moved a small amount. We'll have the length increase by δL >0 . By [Equation 1] this causes the energy of the particle in the box to drop, we have a change in the energy δE.
We'll have a closed system, so that energy can only pass from the particle in the box to/from the walls. Assuming the law of conservation of energy, then any energy lost by the particle must have been work done on the walls.
The work done on the walls should be F.δL = (Force on the wall) x (Distance moved).
So we have FδL = -δE => F = -δE / δL => F =
taking the limit δL -> 0.From [Equation 1] we then see F = k / L3 for k =
= a constant.For a 3-D box we would then have Pressure = P = F / (Area of a wall) = k / L5.
Anyway, that's the basic argument to show that a pressure is exerted on the walls. Arguments for and against can be made just by taking the assumptions of this method apart piece by piece.
Examples: (i) You may have noticed that E as given by [Equation 1] does depend on n, the quantum number of the particle. There is, a priori, no reason why the particle had to start and end with that same quantum numbers for a change in length of the box. Take n = 1 and L = 1m initially. Adjust to n=2 and L = 2m finally. From [Equation 1] we will still have initial Energy = final Energy, no work had to be done on the walls.
You could try and argue that the box had a length = 1.5m and all fractions between 1m and 2m during that change and there was no suitable integer n that could maintain a constant energy for the particle in the box. However, you could have had the particle exert force on the walls for half of the expansion and then retard the walls to recover work from it for the other half of the expansion. While a "negative pressure" is not something we would like in ordinary Newtonian mechanics, it is exactly what we need in theories like General relativity. We cannot explain the current expansion of space without assuming the existence of a component of the cosmological fluid that does exert negative pressure. So you can dismiss complaints about "negative pressure" and just recognise that if QM permits "tunneling" and other strange effects then negative pressure cannot be taken out of the running.
I said "a priori" and we should qualify that a litle: A state does not evolve in time arbitrarily. Unless a measurement is made, the time dependant Schrodinger equation is taken as the only thing which governs the evolution of the state. However, to apply it rigorously, the potential V which the particle experiences will need to be time dependant. As the walls move, the potential barrier is retreating. The full evolution of the state of the particle is going to be complicated and no-one (well no-one I've found from an hours search on the internet) seems to have done the calculation. It could very well be that exactly how you allow the wall to move affects the final state of the particle in the box, i.e. it's not enough to know that you end up with a Length L = 2m, you need to know how you got there.
--- This post is already too long. I'm ending. There are several arguments for and against and I'm sure I don't know half of them ----
Best Wishes.
Post by: Bored chemist on 20/04/2023 17:55:13
Why would using a QM description of it stop that?
Post by: Eternal Student on 20/04/2023 19:40:46
In reality, a particle in a box exerts a pressure on the walls.The model is just a model of one thing. It is not a model of everything that exists or can be exhibited in the universe.
Why would using a QM description of it stop that?
A Newtonian model of a marble rolling down a hill is a perfectly fine model. In reality, the marble was red. The Newtonian description of a sphere rolling down an inclined plane just isn't going to tell you that
For the "particle in a box" model, the walls were not included as a quantum mechanical object. One reasonable argument is that the QM model cannot say much about the walls until you stop treating them as Newtonian objects and instead consider a much more complicated model. For example, replace the walls with a million particles each of which has its own description as a QM object and is included in the wave function of the system.
Best Wishes.
Post by: paul cotter on 20/04/2023 20:31:01
Post by: Eternal Student on 20/04/2023 21:10:09
what I meant to say is that your original question involves the use of classical mechanics and quantum theory in an interaction-can one do this in a rigorous manner?Well yes and no.
No --> There are lots of replies (especially from me) already stating that at some point a jump from Quantum mechanics to Newtonian mechanics is required. This is dangerous and reduces the validity. There has been one very recent reply suggesting that perhaps you do need to start modelling the wall as a quantum mechanical object in its own right etc.
Yes ---> Almost exactly this question, much as it appeared in the poll, can be asked in undergraduate lectures about Physics. Usually it's just a discussion but sometimes it's homework, or even a multiple choice exam question.
See: https://www.physicsforums.com/threads/force-exerted-by-a-particle-in-a-box-on-the-boundary.935415/
for one example from 2017.
What I mean is that the sense of the question and the validity of the answer is sufficiently well accepted that it could even be a multiple choice question like the one the poor lady was facing in 2017 - there is no room for any discussion in a multiple choice question.
More generally, a whole bucket of results in thermodynamics and statistical mechanics are based on starting with a Quantum Mechanical model and quickly passing over to a macroscopic effect or property that can be measured. This is usually done by allowing an expected value of a measurement of a QM system (i.e. an average as if the measurement was done many times) to be regarded as a macroscopic property that would be present in classical physics.
Overall QM is just too complicated to model everything with it and "bridges" to macroscopic and classical physics are required all the time. In a forum like this, we at least have some time and space to discuss the limitations of the model and the bridges that were taken to reach a macroscopic effect.
Best Wishes.
Post by: alancalverd on 20/04/2023 21:56:15
The energy that a particle in a box can have is given by:And there is the weakness in the argument - inadequate specification.
These are the only possible states for an ideal particle making elastic collisions with the sides of a rigid box. As soon as you introduce the possibility of movement by δL you have broken the boundary conditions.
That's where the quantum model needs to be replaced by the classical continuum if you want to derive radiation pressure of a macroscopic system by differentiating energy density with respect to length. As long as the results are consistent, no problem - and they are.
Post by: Eternal Student on 23/04/2023 04:20:31
* Poll closes in a few hours.
* Open to guests.
* Members can alter their selection. I wouldn't bother - but I think I did tick that option when I set it up.
There is no great reward but equally no penalty for voting. Thank you to everyone who has spent any time here.
Best Wishes.
Post by: Bored chemist on 23/04/2023 11:15:40
QM model cannot say much about the wallsNope,
But observation can.
A particle in a box is observed to exert a pressure on the walls.
If QM says it doesn't then it's not reality which is wrong.
Post by: alancalverd on 23/04/2023 15:37:53
For the "particle in a box" model, the walls were not included as a quantum mechanical object.Neither was the particle! The idealised classical model led to the concept of quantised energy levels. Cart/horse inversion strikes again?
Once you have established the notion of the probability wave function, you can consider the possibility of a nonzero value inside the boundary of a real wall, and the concept of tunneling makes sense, but it's not a good starting point.
Post by: Eternal Student on 23/04/2023 23:03:03
A particle in a box is observed to exert a pressure on the walls.Yes, I would have thought so BUT I haven't tried it.
Trying to get one small particle, say an electron, into a box is tricky. It's even harder if you want to make sure it isn't just absorbed into the walls and I'll guess you need very delicate equipment to measure a tiny force on the walls.
Neither was the particle! The idealised classical model led to the concept of quantised energy levels. Cart/horse inversion strikes again?Yes. Cart/horse inversion. Historically, certain things lead toward the development of modern Quantum mechanics but modern QM is a self-contained and hopefully consistent system now. It does not need to maintain any of those pre-1925 assumptions, many of those assumptions were dropped and NOT carried forward into the postulates of modern QM. It is perfectly reasonable to start from the postulates of modern QM and work outwards from there. The modern "particle in a box" model is something that can be derived using modern QM. This is only 6 postulates (depending on how you write them down but you can do it sensibly in 6 short postulates), none of which mention anything like rigid walls, elastic collisions or the QM object acting like an idealised particle (see your sentence which is quoted next).
These are the only possible states for an ideal particle making elastic collisions with the sides of a rigid box.I've noticed a few posts you've made that seem to be based on some belief that historical and discontinued theories and models must still be adhered to. It is certainly interesting to consider the history and perhaps that is the "bridge" that you would choose to take. By the term "bridge" I mean your preferred method to link the modern QM "particle in a box" model to much more macroscopic and classical phenomena but it is not the only bridge or option available and it does not make your choice of bridge more correct or absolute.
Modern QM is under no obligation to maintain all the assumptions of some theory which preceded it.
Best Wishes.
Post by: alancalverd on 24/04/2023 10:47:33
Planck's model was just that - a model from which you can derive useful hypotheses such as the quantisation of electron energy levels in an atom, and the black body spectrum, which stand up to experimental investigation. It is not a "discontinued theory" as it doesn't actually imply that these phenomena are due to particles rattling about in boxes, nor does it make any assumptions: it is merely a mathematical analysis of an ideal particle in an ideal box.
And it probably wouldn't apply to an electron in a real box. One of the problems we have with designing x-ray tubes is that electrons, being charged, tend to stick to the sides instead of making nice elastic collisions. But the x-ray spectrum is pretty much as Planck predicts.
A lot of quantum mechanics actually derives from the impossibility of classical assumptions, in particular the Bohr atom with orbiting electrons, which is inherently unstable. Is it still taught in schools, I wonder? I did complain when it appeared in the Warwick University Coat of Arms, granted in 1967, about 50 years after it had been abandoned as ludicrous.
Post by: Eternal Student on 24/04/2023 21:01:45
Well the poll has closed so it's time to do something with the data.
However, we're just going to leave everyone on the cliff edge - because that's where we have the greatest potential.
ba..da...da... (on the drums)
Best Wishes.